If a + b + c = 1 and a, b, c are all distinct positive reals, then prove that (1- a) (1 - b) (1 - c) > 8 abc.

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Solution

a, b, c are positive reals
Apply AM > GM for two by two
$a + b > 2 \sqrt{a b}$
$b + c > 2 \sqrt{b c}$
$c + a > 2 \sqrt{c a}$
Since $a + b + c = 1$ we have
$1 - c > 2 \sqrt{a b}$
$1 - a > 2 \sqrt{b c}$
$1 - b > 2 \sqrt{c a}$
$(1 - a) (1 - b) (1 - c) > 8 \sqrt{a b} \sqrt{b c} \sqrt{c a}$
$\Rightarrow (1 - a) (1 - b) (1 - c) > 8 a b c$

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